According to method of excitation D.C. generators are classified as follows:

1. Separately excited generators,

2. Self-excited generators.

**3.1.1. Separately excited generators **

These are those generators whose field magnets are energised from an independent external source of D.C. current. Such a generator is shown in Fig. 15.

**3.1.2. Self-excited generators **

These are those generators whose field magnets are energised by the current produced by the generators themselves. Due to residual magnetism, there is always present some flux in the poles. When the armature is rotated, some e.m.f. and hence some induced current produced which is partly or fully passed through the field coils thereby strengthening the pole flux.

Self excited generators can be divided, in accordance with how the field winding is connected into generators, as follows:

(i) Shunt wound generators

(ii) Series wound generators

(iii) Compound wound generators:

- Short shunt
- Long shunt

i. **Shunt wound generators**: Refer Fig. 16. In these generators the field windings are connected across or in parallel with the armature conductors, and have the full voltage of the generator across them.

**Important relations**: Refer Fig. 16.

(iv) Power developed = E_{g}I_{a}

(v) Power delivered = VI

where I_{sh} = shunt field current;

I (or I_{l}) = load current,

I_{a} = armature current,

R_{a} = armature resistance,

E_{g} = generated e.m.f.

Rsh = shunt field resistance, and

V = terminal voltage.

**Example ****8**. A shunt generator supplied 500 A at 500 V. Calculate its generated e.m.f. if its. armature and shunt field resistances are 0.02 Ω and 125 n respectively.

**Solution**. Load current, I = 500 A

Terminal voltage, V = 500 volts

Armature resistance, R_{a} = 0.2 Ω

Shunt field resistance, R_{sh} = 125 Ω

Refer Fig. 21.

Shunt field current,

**Example 9**. A 4-pole, D.C. shunt generator, with a shunt field resistance of 100 ohms and an armature resistance of 1 ohm, has 378 wave-connected conductors in its armature. The flux per pole is 0.02 Wb. If a load resistance of 10 ohms is connected across the armature terminals and the generator is driven at 1000 r.p.m., calculate power absorbed by load.

**Solution**. Given: p = 4 ; R_{sh} = 100 Ω ; R_{a}= 1 Ω, Z = 378 ; a = 2 ;

Ф = 0.02 Wb ; R_{L} = 10 Ω ; N = 1000 r.p.m.

**Power absorbed by the load: **

The generator arrangement is shown in the Fig. 22.

For the generator,

**Example 10.** A 4-pole lap wound shunt generator supplies to 50 lamps of 100 watts, 200 V each. The field and armature resistances are 50 Ω and 0.2 Ω respectively. Allowing a brush drop of 1 V each brush, calculate the following:

- i. Armature current.
- ii. Current per path.
- iii. Generated e.m.f.
- iv. Power output of D.C. armature.

**Solution**. Number of poles, p = 4

Total lamp load, P = 50 × 100 = 5000 W

Terminal voltage, V = 200 Volts

Field resistance, R_{sh} = 50 Ω

Armature resistance, R_{a} = 0.2 Ω

Voltage drop/brush = 1 V

Refer Fig. 23.

**= ****7.25 ****A. ****(Ans.) **

(iii) **Generated e.m.f., ****E _{g} **

**:**

E_{g} = V + I_{a}R_{a} + brush drop = 200 + 29 × 0.2 + 2 ×1 = 207.8 V

Hence, **generated e.m.f. ****= ****207.8 V. (Ans.) **

(iv) **Power output of D.C. armature: **

Power output of D.C. armature

**Example ****11**. A A-pole, 500 V wave-wound shunt generator delivers a load current of 140 A. It has 65 slots with 12 conductors / slot and runs at 800 r.p.m. The shunt field and armature resistances are 250 Ω and 0.2 Ω respectively. The diameter of the bore of the pole shoe is 45 cm, the angle subtended by the pole shoe is 70^{0} and it is 25 cm in length. Assuming contact drop / brush as 1 V, calculate the flux density in the air gap.

**Solution**. Refer Fig. 24.

Number of poles, p = 4

Number of parallel paths, a = 2 [generator being wave wound]

Terminal voltage, V = 500 Volts

Load current, I= 140 A

Speed of rotation, N = 800 r.p.m.

Number of slots = 65

Number of conductors/slot = 12

Total number of conductors, Z = 12 × 65 = 780

Fig. 24

Shunt field resistance, R_{sh} = 250 Ω

Armature resistance, R_{a} = 0.2 Ω

Diameter of the bore of pole shoe, D = 45 cm (= 0.45 m)

Angle subtended by the pole shoe, Ф= 70°

Length of pole shoe = 25 cm (= 0.25 m)

Contact drop/brush = 1 V

**Flux density in the air gap, ****B: **

Area of-pole shoe, A = Arc length × length of pole shoe

= 0.275 × 0.25 = 0.06875 m^{2}

Hence, **flux density in the air gap ****= ****0.371 T. (Ans.)**

**Example 12**. A series generator delivers a current of 100 A at 250 V. Its armature and series field resistances are 0.1 Ω and 0.055 Ω respectively. Find:

(i) Armature current

(ii) Generated e.m.f.

**Solution**. See Fig. 25.

Load current, I= 100 A

Terminal voltage, V = 250 Volts

Armature resistance, R_{a}=0.1 Ω

Series field resistance, R_{se} = 0.055 Ω

(i) **Armature current, ****I _{a} **

**:**

Armature current (I_{a}) = load current (I)

I_{a} =** ****100 A. (Ans.) **

(ii) **Generated e.m.f. : **

Generated e.m.f. E_{g} = V + I(R_{a} + R_{sc})

= 250 + 100(0.1 + 0.055) = 265.5 V

Hence, **generated ****e.m.f, ****= ****265.5 V. ****(Ans.) **

**Example 13**. A short shunt compound generator has armature, series field and shunt field resistances of 0.06 Ω, 0.03 Ω and 110 Ω respectively. It supplies 100 lamps rated at 250 V, 40 W. Find the generated e.m.f. Assume that contact drop / brush = 1 V.

**Solution**. See Fig. 26.

Armature resistance, R_{a} = 0.06 Ω

Series field resistance, R_{se} = 0.03 Ω

Shunt field resistance, R_{sh} = 110 Ω

Terminal voltage, V = 250 Volts

Lamp load, P = 100 × 40 = 4000 W

Contact drop/brush = 1 V

**Generated e.m.f., Eg : **

Voltage drop in series winding

= IR_{se} = 16 × 0.03 = 0.48 V

Voltage across shunt field winding

= V + IR_{se} = 250 + 0.48 = 250.48 V

Armature current, I_{a} = I + I_{sh}

= 16 + 2.277 = 18.277 A

Generated e.m.f., E_{g} = V + IR_{se} + I_{a}R_{a} + brush drop

= 250 + 0.48 + 18.277 × 0.06 + 2 × 1

**= ****253.58 V. ****(Ans.) **

**Example ****14.** A long shunt compound generator has an armature, series field and shunt field resistances of 0.04 Ω, 0.03 Ω and 200 Ω respectively. It supplies a load current of 180 A at 400 V. Calculate the generated e.m.f. Assume contact drop/brush = 1 V.

**Solution**. Refer Fig. 27.

Armature resistance, R_{a} = 0.04 Ω

Series field resistance, R_{se} = 0.03 Ω

Shunt field resistance, R_{sh} = 200 Ω

Load current, I= 180 A

Terminal voltage, V = 400 Volts

Contact drop/brush = 1 V

**Generated e.m.f., E _{g}: **

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